## Maharashtra State Board Class 8 Maths Solutions Chapter 7 Variation Practice Set 7.1

Question 1.

Write the following statements using the symbol of variation.

- Circumference (c) of a circle is directly proportional to its radius (r).
- Consumption of petrol (l) in a car and distance traveled by that car (d) are in direct variation.

Solution:

- c ∝ r
- l ∝ d

Question 2.

Complete the following table considering that the cost of apples and their number are in direct variation.

Number of apples (x) |
1 | 4 | __ | 12 | __ |

Cost of apples (y) |
8 | 32 | 56 | __ | 160 |

Solution:

The cost of apples (y) and their number (x) are in direct variation.

∴y ∝ x

∴y = kx …(i)

where k is the constant of variation

i. When, x = 1, y = 8

∴ Substituting, x = 1 and y = 8 in (i), we get y = kx

∴ 8 = k × 1

∴ k = 8

Substituting k = 8 in (i), we get

y = kx

∴ y = 8x …(ii)

This the equation of variation

ii. When,y = 56, x = ?

∴ Substituting y = 56 in (ii), we get

y = 8x

∴ 56 = 8x

∴ x = \(\frac { 56 }{ 8 }\)

∴ x = 7

iii. When, x = 12, y = ?

∴ Substituting x = 12 in (ii), we get

y = 8x

∴ y = 8 × 12

∴ y = 96

iv. When, y = 160, x = ?

∴ Substituting y = 160 in (ii), we get

y = 8x

∴ 160 = 8x

∴ x = \(\frac { 160 }{ 8 }\)

∴ x = 20

Number of apples (x) |
1 | 4 | 7 | 12 | 20 |

Cost of apples (y) |
8 | 32 | 56 | 96 | 160 |

Question 3.

If m ∝ n and when m = 154, n = 7. Find the value of m, when n = 14.

Solution:

Given that,

m ∝ n

∴ m = kn …(i)

where k is constant of variation.

When m = 154, n = 7

∴ Substituting m = 154 and n = 7 in (i), we get

m = kn

∴ 154 = k × 7

∴ \(k=\frac { 154 }{ 7 }\)

∴ k = 22

Substituting k = 22 in (i), we get

m = kn

∴ m = 22n …(ii)

This is the equation of variation.

When n = 14, m = ?

∴ Substituting n = 14 in (ii), we get

m = 22n

∴ m = 22 × 14

∴ m = 308

Question 4.

If n varies directly as m, complete the following table.

m |
3 | 5 | 6.5 | __ | 1.25 |

n |
12 | 20 | __ | 28 | __ |

Solution:

Given, n varies directly as m

∴ n ∝ m

∴ n = km …(i)

where, k is the constant of variation

i. When m = 3, n = 12

∴ Substituting m = 3 and n = 12 in (i), we get

n = km

∴ 12 = k × 3

∴ \(k=\frac { 12 }{ 3 }\)

∴ k = 4

Substituting, k = 4 in (i), we get

n = km

∴ n = 4m …(ii)

This is the equation of variation.

ii. When m = 6.5, n = ?

∴ Substituting, m = 6.5 in (ii), we get

n = 4m

∴ n = 4 × 6.5

∴ n = 26

iii. When n = 28, m = ?

∴ Substituting, n = 28 in (ii), we get

n = 4m

∴ 28 = 4m

∴ 28 = 4m

∴ \(m=\frac { 28 }{ 4 }\)

∴ m = 7

iv. When m = 1.25, n = ?

∴ Substituting m = 1.25 in (ii), we get

n = 4m

∴ n = 4 × 1.25

∴ n = 5

m |
3 | 5 | 6.5 | 7 | 1.25 |

n |
12 | 20 | 26 | 28 | 5 |

Question 5.

y varies directly as square root of x. When x = 16, y = 24. Find the constant of variation and equation of variation.

Solution:

Given, y varies directly as square root of x.

∴ y ∝ √4x

∴ y = k √x …(i)

where, k is the constant of variation.

When x = 16 ,y = 24.

∴ Substituting, x = 16 and y = 24 in (i), we get

y = k√x

∴24 = k√16

∴24 = 4k

∴ \(k=\frac { 24 }{ 4 }\)

∴ k = 6

Substituting k = 6 in (i), we get

y = k√x

∴ y = 6√x

This is the equation of variation

∴ The constant of variation is 6 and the equation of variation is y = 6√x .

Question 6.

The total remuneration paid to laborers, employed to harvest soybean is in direct variation with the number of laborers. If remuneration of 4 laborers is Rs 1000, find the remuneration of 17 laborers.

Solution:

Let, m represent total remuneration paid to laborers and n represent number of laborers employed to harvest soybean.

Since, the total remuneration paid to laborers, is in direct variation with the number of laborers.

∴ m ∝ n

∴ m = kn …(i)

where, k = constant of variation

Remuneration of 4 laborers is Rs 1000.

i. e., when n = 4, m = Rs 1000

∴ Substituting, n = 4 and m = 1000 in (i), we get m = kn

∴ 1000 = k × 4

∴ \(k=\frac { 1000 }{ 4 }\)

∴ k = 250

Substituting, k = 250 in (i), we get

m = kn

∴ m = 250 n …(ii)

This is the equation of variation

Now, we have to find remuneration of 17 laborers.

i. e., when n = 17, m = ?

∴ Substituting n = 17 in (ii), we get

m = 250 n

∴ m = 250 × 17

∴ m = 4250

∴ The remuneration of 17 laborers is Rs 4250.

**Maharashtra Board Class 8 Maths Chapter 7 Variation Practice Set 7.1 Intext Questions and Activities**

Question 1.

If the rate of notebooks is Rs 240 per dozen, what is the cost of 3 notebooks?

Also find the cost of 9 notebooks, 24 notebooks and 50 notebooks and complete the following table. (Textbook pg. no. 35)

Number of notebooks (x) |
12 | 3 | 9 | 24 | 50 | 1 |

Cost (In Rupees) (y) |
240 | __ | __ | __ | __ | 20 |

Solution:

As the number of notebooks increases their cost also increases.

∴ Number of notebooks and cost of notebooks are in direct proportion.

i.

∴ y = 3 × 20

∴ y = 60

ii.

∴ y = 9 × 20

∴ y = 180

iii.

∴ y = 24 × 20

∴ y = 480

iv.

∴ y = 50 × 20

∴ y = 1000

Number of notebooks (x) |
12 | 3 | 9 | 24 | 50 | 1 |

Cost (In Rupees) (y) |
240 | 60 | 180 | 480 | 1000 | 20 |